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In my real analysis text book there is a question that says:

Decide whether $f(x)=[x]$ is bounded above or below on the interval $[0,a]$ where $a$ is arbitrary, and whether the function takes on it's maximum or minimum value within that same interval.

This question is very straightforward, assuming $[x]=x$. But if that is the case, then the choice of notation is very strange.

Is there another way to interpret the notation's meaning?

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I have seen $\[x\]$ denote the fractional part of $x$, but that doesn't seem to be the desired meaning in this context. –  Brandon Carter May 1 '11 at 22:44
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I've seen that notation to be the nearest integer function before. –  yunone May 1 '11 at 22:44
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Could you specify what textbook you're using? I'm guessing there should be a place where the notation used is explained... –  Guess who it is. May 1 '11 at 22:47
    
[x] is often used as the Iverson Bracket, 1 if is x is true, though that's clearly not the usage here. I've also seen it used for either floor or ceiling, when the right brackets (missing either the top or bottom) were difficult to produce. –  wnoise May 1 '11 at 23:00
    
The question is from Michael Spivak's "Calculus" - 4th Edition (also 3rd edition). It's part (xii) of chapter 7 question 1. –  objectivesea May 1 '11 at 23:20

1 Answer 1

up vote 7 down vote accepted

It had been fairly standard for $[x]$ to represent "the greatest integer not greater than $x$" (aka, the "floor" function). With fancier type-setting options allowing for $\lfloor x \rfloor$ for a more suggestive "floor" notation ---as well as $\lceil x \rceil$ for the counterpart "ceiling" ("smallest integer not smaller than $x$")--- I've seen $[x]$ taking on the role of "nearest integer" (that is, the "rounding" function) although $\lfloor x \rceil$ is also available for this, freeing up $[x]$ for author's discretion.

With regard to @Brandon's "fractional part", I've seen that more often represented as $\{ x \}$, usually in conjunction with the floor interpretation of $[x]$, so that one would write $x = [x] + \{x\}$ (at least for non-negative $x$).

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